A polygon with 9 sides and 9 internal angles are called as Nonagon.
Nonagons are also known as Enneagon.

## Regular Nonagon

When all sides of the Nonagon are of equal length angles with it is called as Regular Nonagon. Normally, regular nonagon has an internal angle of 140o.
In the regular nonagon, the measure of all the angles is also equal.

## Irregular Nonagon

When sides if the given Nonagon are of unequal lengths, it is irregular nonagon.
The interior angles are also unequal.

## Convex Nonagon

The nonagon whose all interior angles measure less than 180o is called as convex nonagon.
All 9 vertices of this nonagon will point outwards.

## Concave Nonagon

If one or more than one interior angle of the nonagon is a reflex angle (angle greater than 180o), it is known as Concave nonagon.
In concave nonagon one or more vertices may point inwards.

In the above nonagon, two vertices are pointing inwards, so there are two reflex interior angles.

## Area of a Nonagon

We can calculate area of the regular nonagon by applying three different methods.

1)    If the length of the side of the nonagon is known.

Consider one of the triangles .Draw the altitude (perpendicular ) from the apex  to the base of the triangle.(The side of the nonagon forms the base.) Let the height of the altitude be h, base of the triangle be s .

The angle at the apex  = $\frac{360^o}{n}$, where n: side of the nonagon  = 9

Angle at apex =$\frac{360^o}{9}$ =400       Since t = $\frac{1}{2}$ the angle at vertex t=200.

Consider tan ratio;

tan t = $\frac{(s/2)}{h}$

Solving for h,

tan t = $\frac{(s/2)}{h}$ = $\frac{s}{2h}$

h= $\frac{s}{2}$ tan t

Area of the triangle = $\frac{1}{2}$ $\times base \times height$
= $\frac{1}{2} \times s \times h$

Consider h in terms of s,

Area = $\frac{1}{2}$$\times s \times (\frac{s}{2tan t}) = \frac{s^2}{(4 tan t)} Since for the regular nonagon , t =20o, tan t =tan 20o = 0.364 So, area = \frac{s^2}{4} \times 0.364 = 0.6869 \times s2 As there are 9 such triangles in a nonagon, the total area of the nonagon = 9 \times area of the triangle. Area of nonagon = 9 \times 0.6869 \times s2 =6.1818 \times s2. 2) When Radius r is known: The radius of the regular polygon is the distance between the center and the vertex of the polygon (radius of the circumcircle). In the triangle, the side of the isosceles triangle is the radius. Draw the perpendicular h (height of the triangle.) Let x be the half of the base of the triangle. Let us write x and h in terms of r. sin t = \frac{x}{r} So , x = r sin t cos t = \frac{h}{r} h = r cos t Area of the triangle = \frac{1}{2}$$\times base \times height$
=x $\times$ h      (since x = $\frac{1}{2}$ base)

In terms of r,

A = r sin t $\times$ r cos t

= $\frac{1}{2}$ r2 sin(2t)

{Use double angle formula for sin : sin (2t) = 2 sin t $\times$ cos t}

Total area for the nonagon  = 9 $\times$ area of the triangle

= 9 $\times$ r2  $\times$ $\frac{sin (2t)}{2}$

Plugging t = $\frac{1800}{n}$ = $\frac{1800}{9}$ = 200

2t =400

sin 40 = 0.6428

Area =  9 $\times$ $\frac{r^2}{2}$ $\times$ 0.6428

A =    2.8925 r2

3)    When the apothem (radius of the incircle) is given

Apothem of the nonagon is the perpendicular distance from the center to the side.

Let a be the length of the apothem, s be the side.

Area of the triangle = $\frac{!}{2}$ $\times$ a $\times$ s

## Exterior Angles of a Nonagon

For exterior angle, use the formula $\frac{360^o}{n}$, where n is the number of sides of the polygon.
Plug in n=9, exterior angle = $\frac{360^o}{9}$ = 40o

## Nonagon Solved Examples

### Solved Examples

Question 1: Compute the sum of all interior angles of the nonagon.
Solution:

Solution: Sum of interior angles of a polygon = (n-2) $\times$ 180o

plug in n=9,

Sum = (9-2) 180 =7 $\times$ 180 =1260o

Sum of interior angles of the nonagon is 1260o

Question 2: Find the area of the regular nonagon whose side measures 6 cm. Round off the answer to nearest hunderdths.
Solution:

Formula for area of nonagon when side is given: = 6.1818 $\times$ s2.

s = length of the side = 6cm

A = 6.1818 $\times$ s2.
= 6.1818 $\times$ (6)2
=222.5448 sq cm.
Area of the nonagon with side length of 6 cm will be 222.54 sq cm.

Question 3: Find the area of a regular nonagon whose radius measures 5 inches.
Solution:

Area of the nonagon with given radius =

A =    2.8925 r2

r = 5 in

A= 2.8925 * 52

= 72.3125 sq in

The area of the nonagon with 5 inches  radius will be 72.3125 sq in.

Question 4: The measure of the exterior angle of a regular polygon is 40o. Determine the number of sides of the polygon.
Solution:

For a regular polygon, all the exterior angles are congruent.

Since total of the exterior angles =360o,  measure of each exterior angle is given by the formula: $\frac{360^o}{n}$

So, $\frac{360^o}{n}$ =40o

Solve for n, n= $\frac{360^o}{40^o}$

n=9

The polygon with exterior angle measuring 40o has 9 sides.